Question:
\[
\int_0^{\pi/2}\frac{\cos2x}{\sin x+\cos x}\,dx=
\]
\[
\int_0^{\pi/2}\frac{\cos2x}{\sin x+\cos x}\,dx=
\]
Show Hint
Factorize \(\cos2x\) as \((\cos x-\sin x)(\cos x+\sin x)\).
Updated On: May 7, 2026
- \(-1\)
- \(0\)
- \(1\)
- \(\frac{\pi}{2}\)
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The Correct Option is B
Solution and Explanation
Step 1: Use: \[ \cos2x=\cos^2x-\sin^2x \] \[ \cos2x=(\cos x-\sin x)(\cos x+\sin x) \]
Step 2: Therefore, \[ \frac{\cos2x}{\sin x+\cos x} = \cos x-\sin x \]
Step 3: \[ \int_0^{\pi/2}(\cos x-\sin x)\,dx \]
Step 4: \[ =[\sin x+\cos x]_0^{\pi/2} \] \[ =(1+0)-(0+1)=0 \] \[ \boxed{0} \]
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